Show that κ(Qn) = λ(Qn) = n for all positive integer n.
Show that k(Qn) =\lambda (Q_n)=n all positive integer n
I want to prove this by induction. so I start with n=1
Base: n=1 then Q1=K2 , which have k (Q 1) =\lambda (Q1)=1 . So the base is good.
Inductive step:
Assume that this work for n=k , meaning for Qk = K2 \times K2 \times \dots \times K2, k times, k (Qk) = lambda (Qk)=k. I need to show that it work for n=k+1 . I know that Q {k+1} = K2 \times K2 \times \dots \times K2, k+1 times, so Q{k+1} has 2^{k+1} vertices and we obtain Q_{k+1}by taking 2 copies of Q_k and connect their vertices pairwise, so there must be at least k+1 bridges. In each bridge, there must be at least one cut vertices, so there are at least k+1 cut vertices in Q{k+1}, So k (Q{k+1}) = lambda (Q{k+1})=k+1
Hence k(Qn)=lambda(Qn)
number thoery
just need 2 answered
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1. Show that S possess two disjoint subsets, the sum of whose
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